Problem: Let $x,$ $y,$ $z$ be real numbers such that $x + y + z = 5$ and $xy + xz + yz = 8.$  Find the largest possible value of $x.$
Solution: Squaring the equation $x + y + z = 5,$ we get
\[x^2 + y^2 + z^2 + 2(xy + xz + yz) = 25.\]Then $x^2 + y^2 + z^2 = 25 - 2 \cdot 8 = 9.$

By Cauchy-Schwarz,
\[(1^2 + 1^2)(y^2 + z^2) \ge (y + z)^2.\]Then $2(9 - x^2) \ge (5 - x)^2,$ which expands as $18 - 2x^2 \ge 25 - 10x + x^2.$  This simplifies to $3x^2 - 10x + 7 \le 0,$ which factors as $(x - 1)(3x - 7) \le 0.$  Hence, $x \le \frac{7}{3}.$

Equality occurs when $y = z = \frac{4}{3},$ so the maximum value of $x$ is $\boxed{\frac{7}{3}}.$